# std::ratio_add

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< cpp‎ | numeric‎ | ratio

Defined in header `<ratio>` |  |
template< class R1, class R2 > using ratio_add = /* see below */; |  | (since C++11)

The alias template `std::ratio_add` denotes the result of adding two exact rational fractions represented by the std::ratio specializations `R1` and `R2`.

The result is a std::ratio specialization std::ratio<U, V>, such that given Num == R1::num * R2::den + R2::num * R1::den and Denom == R1::den * R2::den (computed without arithmetic overflow), `U` is std::ratio<Num, Denom>::num and `V` is std::ratio<Num, Denom>::den.

### Notes

If `U` or `V` is not representable in std::intmax_t, the program is ill-formed. If `Num` or `Denom` is not representable in std::intmax_t, the program is ill-formed unless the implementation yields correct values for `U` and `V`.

The above definition requires that the result of std::ratio_add<R1, R2> be already reduced to lowest terms; for example, std::ratio_add<std::ratio<1, 3>, std::ratio<1, 6>> is the same type as std::ratio<1, 2>.

### Example

Run this code

    #include <iostream>
    #include <ratio>

    int main()
    {
        using two_third = std::ratio<2, 3>;
        using one_sixth = std::ratio<1, 6>;
        using sum = std::ratio_add<two_third, one_sixth>;

        std::cout << "2/3 + 1/6 = " << sum::num << '/' << sum::den << '\n';
    }

Output:

    2/3 + 1/6 = 5/6

### See also

ratio_subtract (C++11) | subtracts two `ratio` objects at compile-time (alias template)

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